Help with Laplace Transformations and 2nd order ODEs

TFM

so split it up:

[tex] \frac{1}{(s + 2)^3} + \frac{4}{(s + 2)^2} [/tex]

[tex] \frac{4}{(s + 2)^2} = 4te^{-2t} [/tex]

okay so for:

[tex] \frac{1}{(s + 2)^3} [/tex]

we need to use:

[tex] t^n = \frac{n!}{s^{n + 1}} [/tex]

so, would this mean:

[tex] \frac{3!}{s^3} = t^2 [/tex]

we have

[tex] \frac{1}{s^3} [/tex]

Is this relevant, or am I going the wrong way for this one/

TFM

sara_87

for 1/(s+2)^3:
laplace of t^2 is 2!/(s^3) so it is 2/s^3
i dont know why you put 1/s^3.
but you have the right idea.
so if the laplace of t^2 is 2!/(s^3), what is the inverse of 1/(s+2)^3 ?

TFM

So:

[tex] t^2 = \frac{2!}{s^3} [/tex]

we have [tex] \frac{1}{(s + 2)^2} [/tex]

so using the Shift, would that be something like:

[tex] \frac{1}{(s + 2)^2} = (t - 2)^2 [/tex]

bvut I think even it is slightly wrong at least because it doesn't deal with the fact that the top value isn't 2!

TFM

sara_87

For 1/(s+1)^3, let's just look at 1/s^3 ; we know that t^2 is 2/s^3 so 1/s^3 will be the inverse of (1/2)(t^2) right?
but, we dont want 1/s^3, we want 1/(s+1)^3, so what do we do (like we did before)?

TFM

First Shft Theorem

[tex] t^2 = 2/s^3 [/tex]

[tex] \frac{1}{2}t^2 = \frac{1}{s^3} [/tex]

using FST

[tex] \frac{1}{2}t^2 = \frac{1}{b^3} [/tex], where b = s + 1

would this make it:

\frac{1}{2}(t+1)^2 = \frac{1}{b^3} [/tex]

???

TFM

sara_87

I dont see any e's in your answer (there should be).

now as we did before, the second fraction is the laplace transform of: 4te^(-2t)
because if you find the laplace of 4t, this will give you 4/s^2 but we dont want s at the bottom, we want (s+2) so thats why we use the first shift theorem and times the 4t by e^(-2t).

now, can you do the same for the 1/(s+1)^3

TFM

Okay so:

so would this mean:

[tex] t^2 would give: \frac{2}{s^3} [/tex]

thus would:

[tex] \frac{2}{(s + 1)^3} be \frac{1}{2}t^2 e^{-t} [/tex]

???

TFM

sara_87

very good!!

TFM

Great

So now:

[tex] Y = \frac{1}{(s + 2)^3} + \frac{4}{(s + 2)^2} [/tex]

this goes to:

[tex] Y = \frac{1}{2}t^2 e^{-t} + 4te^{-2t} [/tex]

So now we put this value of Y back into:

[tex] s^2(Y) - 4 + 4(sY) + 4Y = \frac{1}{s + 2} [/tex]

???

TFM

sara_87

y(t)=(1/2)t^2(e^-t) + 4t(e^-2t)
This is the answer. dont substitute anything into anything.

TFM

Excellent.

So

[tex] y(t) = \frac{1}{2}t^2 e^{-t} + 4te^{-2t} [/tex]

Tomorrow, I'll start (c), if thats okay?

TFM

sara_87

yep thats fine

TFM

Okay, so:

(c):

[tex] y'' + y = sin(t), y_0 = 1, y_0' = 0 [/tex]

L(y(t)) = Y
L(y'(t)) = sL(y(t)) - y(0)
L(y''(t)) = sL(y'(t))-y'(0) = s^2(L(y(t))-sy(0)-y'(0)


there is no y',


[tex] L(y(t)) = Y [/tex]
[tex] L(y'(t)) = sL(y(t)) - 1 [/tex]
[tex] L(y''(t)) = sL(y'(t))-0 = s^2(L(y(t))-sy(0)-y'(0) [/tex]


and for the sin(t)

[tex] sin(\alpha t) = \frac{\alpha}{p^2 + \alpha^2} [/tex]

since alpha = 1:

[tex] sin(t) = \frac{1}{p^2 + 1} [/tex]

Is this okay, I am not quite sure because the lack of y'

TFM

sara_87

That is absolutely right.
what next?

TFM

Okay so:

[tex] L(y(t)) = Y [/tex]

[tex] L(y'(t)) = sL(y(t)) - 1 [/tex]

[tex] L(y''(t)) = s^2(L(y(t))-sy(0) [/tex]

So now I substitute these into the original equation:

[tex] y'' + y = sin(t) [/tex]

[tex] s^2(L(y(t))-sy(0) + Y = \frac{1}{p^2 + 1} [/tex]

Is this okay? Do I substitute y(0) into sy(0), or is there another value/leave it alone?

TFM

sara_87

That's fine
substitute y(0) and also substitue the value L(y(t))

TFM

Okay, so

[tex] s^2(L(y(t))-sy(0) + Y = \frac{1}{p^2 + 1} [/tex]

[tex] s^2(Y)-1 + Y = \frac{1}{p^2 + 1} [/tex]

[tex] Ys^2 + Y - 1 = \frac{1}{p^2 + 1} [/tex]

Now I need to make the Y the subject:

[tex] Ys^2 + Y = \frac{1}{p^2 + 1} + 1 [/tex]

factorise out

[tex] Y(s^2 + 1) = \frac{1}{p^2 + 1} + 1 [/tex]

divide through:

[tex] Y = \frac{\frac{1}{p^2 + 1} + 1}{s^2 + 1} [/tex]

Now I need to find the inverse.

Firstly, split it up into two fractions:

[tex] Y = \frac{\frac{1}{p^2 + 1}}{s^2 + 1} + \frac{1}{s^2 + 1} [/tex]

Does this look okay?

TFM